Difference between revisions of "1995 IMO Problems/Problem 5"
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Let <math>ABCDEF</math> be a convex hexagon with <math>AB=BC=CD</math> and <math>DE=EF=FA</math>, such that <math>\angle BCD=\angle EFA=\frac{\pi}{3}</math>. Suppose <math>G</math> and <math>H</math> are points in the interior of the hexagon such that <math>\angle AGB=\angle DHE=\frac{2\pi}{3}</math>. Prove that <math>AG+GB+GH+DH+HE\ge CF</math>. | Let <math>ABCDEF</math> be a convex hexagon with <math>AB=BC=CD</math> and <math>DE=EF=FA</math>, such that <math>\angle BCD=\angle EFA=\frac{\pi}{3}</math>. Suppose <math>G</math> and <math>H</math> are points in the interior of the hexagon such that <math>\angle AGB=\angle DHE=\frac{2\pi}{3}</math>. Prove that <math>AG+GB+GH+DH+HE\ge CF</math>. | ||
==Solution== | ==Solution== | ||
+ | Draw <math>AE</math> and <math>BD</math> to make equilateral <math>\triangle EFA</math> and <math>\triangle BCD</math>, and draw points <math>I</math> and <math>J</math> such that <math>IA=IB</math>, <math>JD=JE</math>, directed angle <math>\measuredangle IAB=-\measuredangle CDB</math>, and directed angle <math>\measuredangle JDE=-\measuredangle FAE</math> to make equilateral <math>\triangle AIB</math> and <math>\triangle DJE</math>. Notice that <math>G</math> is on the circumcircle of <math>\triangle AIB</math> and <math>H</math> is on the circumcircle of <math>\triangle DJE</math>. By Ptolemy, <math>GA+GB=GI</math> and <math>HD+HE=HJ</math>. So, <cmath>AG+GB+GH+DH+HE=IG+GH+HJ.</cmath> Notice that octagon <math>AIBCDJEF</math> is symmetric about <math>\overline{BE}</math>. So, <math>IG+GH+HJ\ge IJ=CF</math>. --tigerzhang |
Latest revision as of 20:19, 5 July 2020
Problem
Let be a convex hexagon with
and
, such that
. Suppose
and
are points in the interior of the hexagon such that
. Prove that
.
Solution
Draw and
to make equilateral
and
, and draw points
and
such that
,
, directed angle
, and directed angle
to make equilateral
and
. Notice that
is on the circumcircle of
and
is on the circumcircle of
. By Ptolemy,
and
. So,
Notice that octagon
is symmetric about
. So,
. --tigerzhang